Question

Find the number of ways of arranging the letters of the words DANGER, so that no vowel occupies odd place.

Answer 1

Given:

A word "DANGER".

To find:

Find the number of ways of arranging the letters of the words DANGER, so that no vowel occupies odd place.

Solution:

The given word is DANGER.

Number of letters is 6.

Number of vowels is 2 (A and E).

Number of consonants is 4 (D, N, G, R).

As the vowels cannot occupy odd places, they can be arranged in even places.

Two vowels can be arranged in 3 even places in ³P₂ = 6 ways. Rest of the consonants can arrange in the remaining 4 places in 4! ways.

The total number of arrangements is 6 × 4! = 144.

Answer 2

Solution:

Given word => DANGER

Number of Consonants => 4 that are D,N,G,R

Number of Vowels => 2 that are A,E

The condition is that vowel can't occupy odd places so , we will let them occupy the even places.

There are two vowels which can be arranged in 3 even places in,

$^{3} P_{2} = \frac{3!}{(3-2)!} = \frac{3 * 2}{ 1} = 6 \ ways$

Rest of the 4 consonants can occupy remaining 4 places in,

4! = 4 * 3 * 2 * 1 = 24 ways

Total number of arrangements will be = 6 * 24 = 144 ways